An iterative process is a process that repeats, using its own result value as the start value for the next repeat of the process. Quadratic and cubic equations can be rearranged to allow them to be solved using an iterative process.
Rearrange the equation to have the highest value of `x` on one side of the equation, and all other values on the other side. Square or cube root (as required) both sides of the equation such that one side of the equation has an `x` value only.
With a starting value, evaluate the equation. This will be the starting value for the next iteration e.g.:
`x^2 -3x - 10 = 0`
Rearrange to `x^2 = 3x + 10`
Obtain x on one side:`x = sqrt(3x + 10)`
The iteration is: `x_(n+1) = sqrt(3x_n + 10)`
1. Starting with an initial value of 10, find the solution to `x^2 -3x -10 = 0`.
Rearrange to `x = sqrt(3x + 10)`
Iteration is `x_(n+1) = sqrt(3x_n + 10)`
First value is 10: `x_1 = sqrt(3 xx 10 + 10) = 5.83095`
Substitute new value into the iteration: `x_2 = sqrt(3 x 5.83095 + 10) = 5.24336`
Obtain `x_3 = 5.07248` and repeat until `x_7 = 5.00018`
To three decimal places, the answer is 5.000.
SUbstitute into the original equation to check.
2. Find a positive solution to `x^3 - 4x^2 - 11x +30 = 0`. Use an initial value of 10.
Rearrange: `x^3 = 4x^2 + 11x - 30`
One side equal to x: `x = ^3sqrt(4x^2 + 11x - 30)`
Iteration is `x_1 = ^3sqrt(4x_0^2 + 11x_0 - 30)`
First value is 10: `x_1 = ^3sqrt(4 xx (10)^2 + 11 xx 10 - 30) = 7.82974`
Substitute the new value: `x_2 = ^3sqrt(4 xx (7.82974)^2 + 11 xx 7.82974 - 30) = 6.70433`
and repeat until `x_12 = 5.02828`, `x` = 5.0 to 1dp